# What reaction takes place here? [duplicate]

Upon reading the given question, I realized that:

The first reaction is an aldol condensation reaction between the given compound and formaldehyde.

So the product after 1st step is:

Now, the second part has thrown me off. The current ketone has no $$\alpha$$ hydrogen left. Condensation reactions are out of the question. Cannizzaro reaction can not take place as this is a ketone (and not an aldehyde) also the medium is now acidic. However, I saw one possibility of formation of hemi-acetals. (I did not think about full acetals, as such a structure was not present in the given options to this question)

Which gave me the following product:

However, the answer provided seems to be not this. The given answer is this:

Can someone give a suggestion that where I want wrong and/or how could I reach the answer and also what was wrong in my thinking as well?

This question is taken from the JEE(A)-2016 [Paper - 2]

• If that ketone part would've been an alcohol, an easy intramolecular reaction would take place - giving me a full acetal as well as the answer...if only. – McSuperbX1 Mar 27 '19 at 13:17
• @Waylander, thanks for your input. Amazing. However, when my hemi-acetal is formed, I am already on step (ii). At step (ii), an acidic medium is provided instead of a strongly basic one that is needed for Cannizzaro reaction. – McSuperbX1 Mar 27 '19 at 14:00
• You have misidentified the intermediate at the end of step i. The formaldehyde/NaOH conditions will deliver the diol (intermediate 9 in the scheme in the linked answer) through aldol followed by crossed Cannizzaro. The conditions for step ii are for forming the acetal only. – Waylander Mar 27 '19 at 14:46
• @Waylander, you're correct. Thanks for the explanation. Everything seems to fall in place now. – McSuperbX1 Mar 27 '19 at 16:13

• Isn't it that Cannizzaro reactions happen only between two aldehydes? Sure that there are no $\alpha$ hydrogen present, but they aren't aldehydes either. – McSuperbX1 Mar 27 '19 at 13:48